$\displaystyle \int tan^{-1}u du $

is equal to

$\displaystyle utan^{-1}u-\frac{1}{2} ln(1+u^{2}) $

can somebody explain to me how this is true?

Printable View

- Jan 31st 2008, 03:21 PMakhayoonintegration on inverse trig
$\displaystyle \int tan^{-1}u du $

is equal to

$\displaystyle utan^{-1}u-\frac{1}{2} ln(1+u^{2}) $

can somebody explain to me how this is true? - Jan 31st 2008, 03:50 PMPlato
- Jan 31st 2008, 03:55 PMakhayoon
well thats the thing I haven't been taught these

it's just given to me as a formula...but I would like to understand how it happened so that way I dont have to memorize it...

thanks - Jan 31st 2008, 04:02 PMPlato
- Jan 31st 2008, 04:07 PMakhayoon
well..I really don't care you how you find that notion...all I really want is a hint on how to begin solving that problem really, which is what I always get when I have a question on this forum.

- Jan 31st 2008, 04:46 PMtopsquark
Here's your hint, as Plato suggested:

$\displaystyle \frac{d}{du} \left ( u~tan^{-1}(u)-\frac{1}{2} ln(1+u^{2}) \right )$

$\displaystyle = tan^{-1}(u) + \frac{u}{u^2 + 1} - \frac{1}{2} \left ( \frac{1}{1 + u^2} \cdot 2u \right )$

What does this tell you?

-Dan - Jan 31st 2008, 07:20 PMJhevon
the responses given seem to me to be geared towards verifying that the claim is true, rather than saying why it's true, or how it works. i believe what the poster was after is actually integrating arctan to get the result shown. in that case, we use integration by parts with $\displaystyle u = \arctan x$ and $\displaystyle dv = 1$ (you do know what i am referring to when i say $\displaystyle u \mbox{ and } dv$, right akhayoon?)

- Jan 31st 2008, 07:24 PMKrizalid
A similar integration by parts reasoning is useful here: take the following function $\displaystyle f(u)=u\arctan u.$ Contemplate its derivative and integrate, the rest follows.